Plate theory

Deformation of a thin plate highlighting the displacement, the mid-surface (red) and the normal to the mid-surface (blue)

The Kirchhoff–Love theory is an extension of Euler–Bernoulli beam theory to thin plates. The theory was developed in 1888 by Love[2] using assumptions proposed by Kirchhoff. It is assumed that a mid-surface plane can be used to represent the three-dimensional plate in two-dimensional form.

The following kinematic assumptions are made in this theory:[3]

• straight lines normal to the mid-surface remain straight after deformation
• straight lines normal to the mid-surface remain normal to the mid-surface after deformation
• the thickness of the plate does not change during a deformation.

Displacement field

The Kirchhoff hypothesis implies that the displacement field has the form

${\displaystyle {\begin{aligned}u_{\alpha }(\mathbf {x} )&=u_{\alpha }^{0}(x_{1},x_{2})-x_{3}~{\frac {\partial w^{0}}{\partial x_{\alpha }}}=u_{\alpha }^{0}-x_{3}~w_{,\alpha }^{0}~;~~\alpha =1,2\\u_{3}(\mathbf {x} )&=w^{0}(x_{1},x_{2})\end{aligned}}}$

where ${\displaystyle x_{1}}$ and ${\displaystyle x_{2}}$ are the Cartesian coordinates on the mid-surface of the undeformed plate, ${\displaystyle x_{3}}$ is the coordinate for the thickness direction, ${\displaystyle u_{1}^{0},u_{2}^{0}}$ are the in-plane displacements of the mid-surface, and ${\displaystyle w^{0}}$ is the displacement of the mid-surface in the ${\displaystyle x_{3}}$ direction.

If ${\displaystyle \varphi _{\alpha }}$ are the angles of rotation of the normal to the mid-surface, then in the Kirchhoff–Love theory ${\displaystyle \varphi _{\alpha }=w_{,\alpha }^{0}\,.}$

Displacement of the mid-surface (left) and of a normal (right)

For an isotropic and homogeneous plate, the stress–strain relations are

${\displaystyle {\begin{bmatrix}\sigma _{11}\\\sigma _{22}\\\sigma _{12}\end{bmatrix}}={\cfrac {E}{1-\nu ^{2}}}{\begin{bmatrix}1&\nu &0\\\nu &1&0\\0&0&1-\nu \end{bmatrix}}{\begin{bmatrix}\varepsilon _{11}\\\varepsilon _{22}\\\varepsilon _{12}\end{bmatrix}}\,.}$

The moments corresponding to these stresses are

${\displaystyle {\begin{bmatrix}M_{11}\\M_{22}\\M_{12}\end{bmatrix}}=-{\cfrac {2h^{3}E}{3(1-\nu ^{2})}}~{\begin{bmatrix}1&\nu &0\\\nu &1&0\\0&0&1-\nu \end{bmatrix}}{\begin{bmatrix}w_{,11}^{0}\\w_{,22}^{0}\\w_{,12}^{0}\end{bmatrix}}}$

For an orthotropic plate

${\displaystyle {\begin{bmatrix}C_{11}&C_{12}&C_{13}\\C_{12}&C_{22}&C_{23}\\C_{13}&C_{23}&C_{33}\end{bmatrix}}={\cfrac {1}{1-\nu _{12}\nu _{21}}}{\begin{bmatrix}E_{1}&\nu _{12}E_{2}&0\\\nu _{21}E_{1}&E_{2}&0\\0&0&2G_{12}(1-\nu _{12}\nu _{21})\end{bmatrix}}\,.}$

Therefore,

${\displaystyle {\begin{bmatrix}A_{11}&A_{12}&A_{13}\\A_{21}&A_{22}&A_{23}\\A_{31}&A_{32}&A_{33}\end{bmatrix}}={\cfrac {2h}{1-\nu _{12}\nu _{21}}}{\begin{bmatrix}E_{1}&\nu _{12}E_{2}&0\\\nu _{21}E_{1}&E_{2}&0\\0&0&2G_{12}(1-\nu _{12}\nu _{21})\end{bmatrix}}}$

and

${\displaystyle {\begin{bmatrix}D_{11}&D_{12}&D_{13}\\D_{21}&D_{22}&D_{23}\\D_{31}&D_{32}&D_{33}\end{bmatrix}}={\cfrac {2h^{3}}{3(1-\nu _{12}\nu _{21})}}{\begin{bmatrix}E_{1}&\nu _{12}E_{2}&0\\\nu _{21}E_{1}&E_{2}&0\\0&0&2G_{12}(1-\nu _{12}\nu _{21})\end{bmatrix}}\,.}$

The dynamic theory of plates determines the propagation of waves in the plates, and the study of standing waves and vibration modes.

In the theory of thick plates, or theory of Yakov S. Uflyand[4] (see, for details, Elishakoff's handbook[5]), Raymond Mindlin[6] and Eric Reissner, the normal to the mid-surface remains straight but not necessarily perpendicular to the mid-surface. If ${\displaystyle \varphi _{1}}$ and ${\displaystyle \varphi _{2}}$ designate the angles which the mid-surface makes with the ${\displaystyle x_{3}}$ axis then

${\displaystyle \varphi _{1}\neq w_{,1}~;~~\varphi _{2}\neq w_{,2}}$

Then the Mindlin–Reissner hypothesis implies that

For uniformly thick, homogeneous, and isotropic plates, the stress–strain relations in the plane of the plate are

${\displaystyle {\begin{bmatrix}\sigma _{11}\\\sigma _{22}\\\sigma _{12}\end{bmatrix}}={\cfrac {E}{1-\nu ^{2}}}{\begin{bmatrix}1&\nu &0\\\nu &1&0\\0&0&1-\nu \end{bmatrix}}{\begin{bmatrix}\varepsilon _{11}\\\varepsilon _{22}\\\varepsilon _{12}\end{bmatrix}}\,.}$

where ${\displaystyle E}$ is the Young's modulus, ${\displaystyle \nu }$ is the Poisson's ratio, and ${\displaystyle \varepsilon _{\alpha \beta }}$ are the in-plane strains. The through-the-thickness shear stresses and strains are related by

${\displaystyle \sigma _{31}=2G\varepsilon _{31}\quad {\text{and}}\quad \sigma _{32}=2G\varepsilon _{32}}$

where ${\displaystyle G=E/(2(1+\nu ))}$ is the shear modulus.

In general, exact solutions for cantilever plates using plate theory are quite involved and few exact solutions can be found in the literature. Reissner and Stein[7] provide a simplified theory for cantilever plates that is an improvement over older theories such as Saint-Venant plate theory.

The Reissner-Stein theory assumes a transverse displacement field of the form

${\displaystyle w(x,y)=w_{x}(x)+y\,\theta _{x}(x)\,.}$

The governing equations for the plate then reduce to two coupled ordinary differential equations:

where

${\displaystyle {\begin{aligned}q_{1}(x)&=\int _{-b/2}^{b/2}q(x,y)\,{\text{d}}y~,~~q_{2}(x)=\int _{-b/2}^{b/2}y\,q(x,y)\,{\text{d}}y~,~~n_{1}(x)=\int _{-b/2}^{b/2}n_{x}(x,y)\,{\text{d}}y\\n_{2}(x)&=\int _{-b/2}^{b/2}y\,n_{x}(x,y)\,{\text{d}}y~,~~n_{3}(x)=\int _{-b/2}^{b/2}y^{2}\,n_{x}(x,y)\,{\text{d}}y\,.\end{aligned}}}$

At ${\displaystyle x=0}$, since the beam is clamped, the boundary conditions are

${\displaystyle w(0,y)={\cfrac {dw}{dx}}{\Bigr |}_{x=0}=0\qquad \implies \qquad w_{x}(0)={\cfrac {dw_{x}}{dx}}{\Bigr |}_{x=0}=\theta _{x}(0)={\cfrac {d\theta _{x}}{dx}}{\Bigr |}_{x=0}=0\,.}$

The boundary conditions at ${\displaystyle x=a}$ are

${\displaystyle {\begin{aligned}&bD{\cfrac {d^{3}w_{x}}{dx^{3}}}+n_{1}(x){\cfrac {dw_{x}}{dx}}+n_{2}(x){\cfrac {d\theta _{x}}{dx}}+q_{x1}=0\\&{\frac {b^{3}D}{12}}{\cfrac {d^{3}\theta _{x}}{dx^{3}}}+\left[n_{3}(x)-2bD(1-\nu )\right]{\cfrac {d\theta _{x}}{dx}}+n_{2}(x){\cfrac {dw_{x}}{dx}}+t=0\\&bD{\cfrac {d^{2}w_{x}}{dx^{2}}}+m_{1}=0\quad ,\quad {\frac {b^{3}D}{12}}{\cfrac {d^{2}\theta _{x}}{dx^{2}}}+m_{2}=0\end{aligned}}}$

where

${\displaystyle {\begin{aligned}m_{1}&=\int _{-b/2}^{b/2}m_{x}(y)\,{\text{d}}y~,~~m_{2}=\int _{-b/2}^{b/2}y\,m_{x}(y)\,{\text{d}}y~,~~q_{x1}=\int _{-b/2}^{b/2}q_{x}(y)\,{\text{d}}y\\t&=q_{x2}+m_{3}=\int _{-b/2}^{b/2}y\,q_{x}(y)\,{\text{d}}y+\int _{-b/2}^{b/2}m_{xy}(y)\,{\text{d}}y\,.\end{aligned}}}$

Derivation of Reissner–Stein cantilever plate equations

The strain energy of bending of a thin rectangular plate of uniform thickness ${\displaystyle h}$ is given by ${\displaystyle U={\frac {1}{2}}\int _{0}^{a}\int _{-b/2}^{b/2}D\left\{\left({\frac {\partial ^{2}w}{\partial x^{2}}}+{\frac {\partial ^{2}w}{\partial y^{2}}}\right)^{2}+2(1-\nu )\left[\left({\frac {\partial ^{2}w}{\partial x\partial y}}\right)^{2}-{\frac {\partial ^{2}w}{\partial x^{2}}}{\frac {\partial ^{2}w}{\partial y^{2}}}\right]\right\}{\text{d}}x{\text{d}}y}$ where ${\displaystyle w}$ is the transverse displacement, ${\displaystyle a}$ is the length, ${\displaystyle b}$ is the width, ${\displaystyle \nu }$ is the Poisson's ratio, ${\displaystyle E}$ is the Young's modulus, and ${\displaystyle D={\frac {Eh^{3}}{12(1-\nu )}}.}$ The potential energy of transverse loads ${\displaystyle q(x,y)}$ (per unit length) is ${\displaystyle P_{q}=\int _{0}^{a}\int _{-b/2}^{b/2}q(x,y)\,w(x,y)\,{\text{d}}x{\text{d}}y\,.}$ The potential energy of in-plane loads ${\displaystyle n_{x}(x,y)}$ (per unit width) is ${\displaystyle P_{n}={\frac {1}{2}}\int _{0}^{a}\int _{-b/2}^{b/2}n_{x}(x,y)\,\left({\frac {\partial w}{\partial x}}\right)^{2}\,{\text{d}}x{\text{d}}y\,.}$ The potential energy of tip forces ${\displaystyle q_{x}(y)}$ (per unit width), and bending moments ${\displaystyle m_{x}(y)}$ and ${\displaystyle m_{xy}(y)}$ (per unit width) is ${\displaystyle P_{t}=\int _{-b/2}^{b/2}\left(q_{x}(y)\,w(x,y)-m_{x}(y)\,{\frac {\partial w}{\partial x}}+m_{xy}(y)\,{\frac {\partial w}{\partial y}}\right){\text{d}}x{\text{d}}y\,.}$ A balance of energy requires that the total energy is ${\displaystyle W=U-(P_{q}+P_{n}+P_{t})\,.}$ With the Reissener–Stein assumption for the displacement, we have ${\displaystyle U=\int _{0}^{a}{\frac {bD}{24}}\left[12\left({\cfrac {d^{2}w_{x}}{dx^{2}}}\right)^{2}+b^{2}\left({\cfrac {d^{2}\theta _{x}}{dx^{2}}}\right)^{2}+24(1-\nu )\left({\cfrac {d\theta _{x}}{dx}}\right)^{2}\right]\,{\text{d}}x\,,}$ ${\displaystyle P_{q}=\int _{0}^{a}\left[\left(\int _{-b/2}^{b/2}q(x,y)\,{\text{d}}y\right)w_{x}+\left(\int _{-b/2}^{b/2}yq(x,y)\,{\text{d}}y\right)\theta _{x}\right]\,dx\,,}$ ${\displaystyle {\begin{aligned}P_{n}&={\frac {1}{2}}\int _{0}^{a}\left[\left(\int _{-b/2}^{b/2}n_{x}(x,y)\,{\text{d}}y\right)\left({\cfrac {dw_{x}}{dx}}\right)^{2}+\left(\int _{-b/2}^{b/2}yn_{x}(x,y)\,{\text{d}}y\right){\cfrac {dw_{x}}{dx}}\,{\cfrac {d\theta _{x}}{dx}}\right.\\&\left.\qquad \qquad +\left(\int _{-b/2}^{b/2}y^{2}n_{x}(x,y)\,{\text{d}}y\right)\left({\cfrac {d\theta _{x}}{dx}}\right)^{2}\right]{\text{d}}x\,,\end{aligned}}}$ and ${\displaystyle {\begin{aligned}P_{t}&=\left(\int _{-b/2}^{b/2}q_{x}(y)\,{\text{d}}y\right)w_{x}-\left(\int _{-b/2}^{b/2}m_{x}(y)\,{\text{d}}y\right){\cfrac {dw_{x}}{dx}}+\left[\int _{-b/2}^{b/2}\left(yq_{x}(y)+m_{xy}(y)\right)\,{\text{d}}y\right]\theta _{x}\\&\qquad \qquad -\left(\int _{-b/2}^{b/2}ym_{x}(y)\,{\text{d}}y\right){\cfrac {d\theta _{x}}{dx}}\,.\end{aligned}}}$ Taking the first variation of ${\displaystyle W}$ with respect to ${\displaystyle (w_{x},\theta _{x},x)}$ and setting it to zero gives us the Euler equations ${\displaystyle {\text{(1)}}\qquad bD{\frac {\mathrm {d} ^{4}w_{x}}{\mathrm {d} x^{4}}}=q_{1}(x)-n_{1}(x){\cfrac {d^{2}w_{x}}{dx^{2}}}-{\cfrac {dn_{1}}{dx}}\,{\cfrac {dw_{x}}{dx}}-{\frac {1}{2}}{\cfrac {dn_{2}}{dx}}\,{\cfrac {d\theta _{x}}{dx}}-{\frac {n_{2}(x)}{2}}{\cfrac {d^{2}\theta _{x}}{dx^{2}}}}$ and ${\displaystyle {\text{(2)}}\qquad {\frac {b^{3}D}{12}}\,{\frac {\mathrm {d} ^{4}\theta _{x}}{\mathrm {d} x^{4}}}-2bD(1-\nu ){\cfrac {d^{2}\theta _{x}}{dx^{2}}}=q_{2}(x)-n_{3}(x){\cfrac {d^{2}\theta _{x}}{dx^{2}}}-{\cfrac {dn_{3}}{dx}}\,{\cfrac {d\theta _{x}}{dx}}-{\frac {n_{2}(x)}{2}}\,{\cfrac {d^{2}w_{x}}{dx^{2}}}-{\frac {1}{2}}{\cfrac {dn_{2}}{dx}}\,{\cfrac {dw_{x}}{dx}}}$ where ${\displaystyle {\begin{aligned}q_{1}(x)&=\int _{-b/2}^{b/2}q(x,y)\,{\text{d}}y~,~~q_{2}(x)=\int _{-b/2}^{b/2}y\,q(x,y)\,{\text{d}}y~,~~n_{1}(x)=\int _{-b/2}^{b/2}n_{x}(x,y)\,{\text{d}}y\\n_{2}(x)&=\int _{-b/2}^{b/2}y\,n_{x}(x,y)\,{\text{d}}y~,~~n_{3}(x)=\int _{-b/2}^{b/2}y^{2}\,n_{x}(x,y)\,{\text{d}}y.\end{aligned}}}$ Since the beam is clamped at ${\displaystyle x=0}$, we have ${\displaystyle w(0,y)={\cfrac {dw}{dx}}{\Bigr |}_{x=0}=0\qquad \implies \qquad w_{x}(0)={\cfrac {dw_{x}}{dx}}{\Bigr |}_{x=0}=\theta _{x}(0)={\cfrac {d\theta _{x}}{dx}}{\Bigr |}_{x=0}=0\,.}$ The boundary conditions at ${\displaystyle x=a}$ can be found by integration by parts: ${\displaystyle {\begin{aligned}&bD{\cfrac {d^{3}w_{x}}{dx^{3}}}+n_{1}(x){\cfrac {dw_{x}}{dx}}+n_{2}(x){\cfrac {d\theta _{x}}{dx}}+q_{x1}=0\\&{\frac {b^{3}D}{12}}{\cfrac {d^{3}\theta _{x}}{dx^{3}}}+\left[n_{3}(x)-2bD(1-\nu )\right]{\cfrac {d\theta _{x}}{dx}}+n_{2}(x){\cfrac {dw_{x}}{dx}}+t=0\\&bD{\cfrac {d^{2}w_{x}}{dx^{2}}}+m_{1}=0\quad ,\quad {\frac {b^{3}D}{12}}{\cfrac {d^{2}\theta _{x}}{dx^{2}}}+m_{2}=0\end{aligned}}}$ where ${\displaystyle {\begin{aligned}m_{1}&=\int _{-b/2}^{b/2}m_{x}(y)\,{\text{d}}y~,~~m_{2}=\int _{-b/2}^{b/2}y\,m_{x}(y)\,{\text{d}}y~,~~q_{x1}=\int _{-b/2}^{b/2}q_{x}(y)\,{\text{d}}y\\t&=q_{x2}+m_{3}=\int _{-b/2}^{b/2}y\,q_{x}(y)\,{\text{d}}y+\int _{-b/2}^{b/2}m_{xy}(y)\,{\text{d}}y.\end{aligned}}}$